
A ball of mass 200 gm is moving with velocity of $10m{{s}^{-1}}$. If the error in measurement of velocity is 0.1 %, the uncertainty in its position is:
(A) $3.3\times {{10}^{-31}}m$
(B) $3.3\times {{10}^{-27}}m$
(C) $5.3\times {{10}^{-25}}m$
(D) $2.62\times {{10}^{-32}}m$
Answer
232.8k+ views
Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.
Complete step by step answer
We should know that:
The uncertainty in position is the accuracy of the measurement, or $\Delta x\text{ }=\text{ }0.0100\text{ }nm$. Thus, the smallest uncertainty in momentum $\Delta p$ can be calculated using $\Delta x\Delta p\ge h4\pi \text{ }\Delta \text{ }x\text{ }\Delta \text{ }p\text{ }\ge \text{ }h\text{ }4\text{ }\pi$. Once the uncertainty in momentum $\Delta p$ is found, the uncertainty in velocity can be found from $\Delta p\text{ }=\text{ }m\Delta v$.
Let us first write quantities that are mentioned in the question:
$\mathrm{m}=200 \mathrm{gm}$
The error in the velocity is given as:
$\Delta \mathrm{v}=0.01 \mathrm{m} / \mathrm{s}$
Now the uncertainty of the position is given as:
$\Delta \mathrm{n}=\dfrac{\mathrm{h}}{4 \pi \times \dfrac{1}{5} \times 0.01}=2.62 \times 10^{-32} \mathrm{m}$
Hence, the uncertainty is $2.62\times {{10}^{-32}}m$.
Hence the correct answer is option D.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Complete step by step answer
We should know that:
The uncertainty in position is the accuracy of the measurement, or $\Delta x\text{ }=\text{ }0.0100\text{ }nm$. Thus, the smallest uncertainty in momentum $\Delta p$ can be calculated using $\Delta x\Delta p\ge h4\pi \text{ }\Delta \text{ }x\text{ }\Delta \text{ }p\text{ }\ge \text{ }h\text{ }4\text{ }\pi$. Once the uncertainty in momentum $\Delta p$ is found, the uncertainty in velocity can be found from $\Delta p\text{ }=\text{ }m\Delta v$.
Let us first write quantities that are mentioned in the question:
$\mathrm{m}=200 \mathrm{gm}$
The error in the velocity is given as:
$\Delta \mathrm{v}=0.01 \mathrm{m} / \mathrm{s}$
Now the uncertainty of the position is given as:
$\Delta \mathrm{n}=\dfrac{\mathrm{h}}{4 \pi \times \dfrac{1}{5} \times 0.01}=2.62 \times 10^{-32} \mathrm{m}$
Hence, the uncertainty is $2.62\times {{10}^{-32}}m$.
Hence the correct answer is option D.
Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

